Seunghee Ye Ma 8: Week 5 Oct 28
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1 Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value Theorem Midterm Review 5 2. Proof Techiques Sequeces Series Cotiuity ad Differetiability of Fuctios Mea Value Theorem The Mea Value Theorem is the followig result: Theorem. (Mea Value Theorem. Let f be a cotiuous fuctio o [a, b], which is differetiable o (a, b. The, there exists some value c (a, b such that f (c = f(b f(a b a Ituitively, the Mea Value Theorem is quite trivial. Say we wat to drive to Sa Fracisco, which is 380 miles from Caltech accordig to Google Map. If we start drivig at 8am ad arrive at 2pm, we kow that we were drivig over the speed it at least oce durig the drive. This is exactly what the Mea Value Theorem tells us. Sice the distace travelled is a cotiuous fuctio of time, we kow that there is a poit i time whe our speed was 380/4 >>> speed it. As we ca see from this example, the Mea Value Theorem is usually ot a tough theorem to uderstad. The tricky thig is realizig whe you should try to use it. Roughly speakig, we use the Mea Value Theorem whe we wat to tur the iformatio about a fuctio ito iformatio about its derivative, or vice-versa.. Applicatios of the Mea Value Theorem Example.. Cosider the equatio (x + y = x + y If either x or y is zero, the equatio holds. Also, if x = y ad is odd, the equatio holds. Are there other values of x, y R ad N for which we ca fid solutios for above equatio? Solutio. We will show that the aswer is o. First, we ote the followig lemma we ca prove usig the Mea Value Theorem: Lemma.. Let f be a differetiable fuctio with k distict roots a < a 2 < < a k. The, f has at least k distict roots b < b 2 < < b k such that a < b < a 2 < b 2 < < a k < b k < a k Page of 0
2 Proof. Sice f is differetiable o [a, a 2 ], by the Mea Value Theorem, we ca fid b (a, a 2 such that f (b = f(a 2 f(a a 2 a = 0 0 a 2 a = 0 I other words, b is a root of f. Repeatig this argumet for [a i, a i+ ] for i =,..., k, we fid k distict roots b,..., b k of f with the desired property. How ca we use this lemma to solve our problem? Let s fix a ozero value y R ad N, ad look at the equatio f (x = (x + y x y Note that we are tryig to fid out which values of x satisfy f(x = 0. Curretly we kow that the followig values of x are roots of f(x:. If is eve, the oly root we kow is x = If is odd, we kow that x = 0 ad x = y are roots Ca there be more distict roots? By Lemma. we kow that if f (x has k distict roots, f (x must have at least k distict roots. So let s see how may roots f (x ca have. Differetiatig we obtai Now, if x is a root of f (x, we have f (x = (x + y x 0 = (x + y x ( x = (x + y (2 x = (x + y (3 Now, we have two cases depedig o whether is eve or odd. If is eve, is odd. Hece, (3 is equivalet to x = x + y y = 0 which cotradicts our ozero choice of y. Hece, whe is eve, f (x does ot have a root. Therefore, f (x ca have at most distict root, amely x = 0. If is odd, the is eve. Now (3 is equivalet to x = x + y ±x = x + y y = 0 or y = 2x Agai, sice we chose y to be ozero, f (x has a uique root x = y 2. By Lemma., we colcude that f (x ca have at most 2 distict roots, amely x = 0 ad x = y. Therefore, there ca be o other type of roots to the origial equatio tha the oes listed. Example. showed how we could tur iformatio about our fuctio (specifically, kowledge of where its roots are ito iformatio about the derivative. The Mea Value Theorem ca also be used to tur iformatio about the derivative ito iformatio about the fuctio as we illustrate here: Example.2. Let p(t deote the curret locatio of a particle movig i a oe-dimesioal space. Suppose that p(0 = 0, p( = ad p (0 = p ( = 0. Show that there must be some poit i time i [0, ] where p (t 4. Solutio. We will prove this by cotradictio. Page 2 of 0
3 Suppose for every t [0, ], we have p (t < 4. Where ca we go from here? We have some boudary coditios, i.e. p(0 = 0, p( =, p (0 = p ( = 0, ad oe piece of global iformatio, i.e. p (t < 4. How ca we tur this kowledge of the secod derivative to iformatio about the rest of the fuctio? What does the Mea Value Theorem say about p (t? It tells us that o ay iterval [a, b], we ca fid c (a, b such that p (b p (a = (p (c = p (c b I particular, we kow that for all [a, b] [0, ] the Mea Value Theorem tells us that p (b p (a b a = p (c < 4 where c (a, b [0, ]. I particular, if we set a = 0, b = t ad remember that p (0 = 0, we see that p (t p (0 t 0 = p (t 0 t = p (t < 4 t Hece, p (t < 4t. Similarly, if we let a = t, b = ad remember p ( = 0, we get p ( p ( t ( t = 0 p ( t t = p ( t < 4 t Hece, p ( t < 4t. Great. So we maaged to tur iformatio about the secod derivative to iformatio about the first derivative. Let s preted for the momet that you are back i your high school calculus course ad you kow how to fid atiderivatives. I this situatio, we have a fuctio p(t with the followig properties: p(0 = 0, p( = p (t < 4t for all t (0, p ( t < 4t for all t (0, Well, if p (t < 4t, we ca itegrate to get p(t < 2t 2 + C. Sice p(0 = 0, we see that p(t < 2t 2 t (0, Now, sice p ( t < 4t we kow that p ( t > 4t. Itegratig, we get p( t > 2t 2 + C ad usig p( =, we see that p( t > 2t 2 + t (0, But what happes whe we take a look at t = 2? I our first iequality, we have But i our secod iequality we have p ( ( 2 < 2 = ( p ( 2 > 2 + = Hece, we get ( 2 < p < 2 2 Page 3 of 0
4 which is a cotradictio. Of course, this was assumig we kew how to fid atiderivatives. Let s solve this problem usig just the tools we have available to us. I fact, let s use the Mea Value Theorem oe more time to solve the problem. Earlier, we tured iformatio about the secod derivative ito iformatio about the first derivative. Ca we try the same trick to get iformatio about the origial fuctio, p(x? By the Mea Value Theorem, we kow that for all t [0, ], we ca fid c (0, t such that p(t p(0 t 0 = p(t t = p (c However, we kow that p (c < 4c < 4t for all c (0, t (0,. Hece, we get p(t t = p (c < 4c < 4t p(t < 4t 2 This is t exactly the boud p(t < 2t 2 we obtaied by itegratig. But we ca take this boud ad look at the iterval [t, 2t] [0, ]. By the Mea Value Theore, we ca fid c (t, 2t (0, such that Hece, we get p(2t p(t 2t t = p (c p(2t p(t = t p (c < t 4c < t 4 2t = 8t 2 p(2t < p(t + 8t 2 < 4t 2 + 8t 2 Hece, for all t such that 2t [0, ], we have p(2t < 4t 2 + 8t 2 = 4t 2 ( + 2 Let s do oe more step. If we look at a iterval of the form [2t, 3t] [0, ], we ca fid c (2t, 3t (0, such that p(3t p(2t = p (c 3t 2t Hece, we get p(3t p(2t = t p (c < t 4c < t 4 3t = 2t 2 p(3t < p(2t + 2t 2 < 4t 2 + 8t 2 + 2t 2 Hece, for all t such that 3t (0,, we coclude that p(3t < 4t 2 + 8t 2 + 2t 2 = 4t 2 ( By a simple iductive argumet, we ca coclude that for wheever t (0, for N, we have Now, let t (0,. The, we ca write t = t p(t < 4( t 2 = 2( + t 2 for all N. By above iequality, we have that for all, ( p(t = p t ( 2 t < 2( + = 2t 2 + Sice this is true for all N, we ca take the it as goes to ifiity. The, we obtai p(t < 2t 2 t (0, Page 4 of 0
5 (Techically, what we get is p(t 2t 2 but we kow that above iequality is true from itegratig Similarly, we ca show that p( t 2t 2 +. Hece, by lettig t = 2, we get ( 2 < p < 2 2 which is a cotradictio. 2 Midterm Review 2. Proof Techiques Example 2.. Suppose that a,..., a [0, ]. Show that ( a i Proof. We proceed by iductio. The base case is trivial. For =, we have i= i= a i ( a i = a a = i= Now, assume that the statemet is true for all k. Pick a,..., a + [0, ]. Note that By iductio hypothesis we see that i= i= i= + ( a i = ( a + ( a i i= + ( a i = ( a + ( a i (4 ( a + i= ( a i i= ( ( = a + a i + a + a i a + i= a i i= (5 (6 a i (7 i= + = a i (8 i= Example 2.2. Show that 5 39 is irratioal. Proof. We proceed by cotradictio. Assume that 5 39 is ratioal. The, we ca fid relatively prime itegers p, q such that 5 39 = p q. By raisig both sides to the fifth power we get 39 = p5 q 5 39q 5 = p 5 Page 5 of 0
6 Note that 3 divides 39q 5. Hece, 3 must also divide p 5. However, sice 3 is a prime umber, it meas that 3 divides p: oe way to defie prime umbers is to defie them as umbers p with the property that if p xy, the p x or p y. This is also ot very hard to prove by cotradictio. Now that we kow 3 p, we ca write p = 3r. Hece, we have 39q 5 = (3r 5 = 3 5 r 5 3q 5 = 3 4 r 5 Sice r 5 = 3q 5, either 3 3 or 3 q 5. We kow that 3 3 ad thus we coclude that 3 q 5. As before, this meas that 3 q. However, ote that 3 p ad 3 q which cotradicts our iitial assumptio that p ad q are relatively prime. Therefore, 5 39 is irratioal. 2.2 Sequeces Example 2.3. Let x, y be a pair of positive real umbers such that x < y. Show that (x + y / = y Proof. We will use the squeeze theorem to solve the problem. Note that But we kow that ad Hece, by the squeeze theorem, we coclude that y (x + y / (y + y / = 2 / y y = y 2/ y = y 2/ = y (x + y / = y Example 2.4. Cosider the followig sequece: a = ( l= k l 2 k=0 Determie whether {a } coverges ad fid the it if it exists. Proof. So this is a pretty crazy lookig sequece. But there s othig to be afraid of. Recall that if f(x is cotiuous at a, ad we have a sequece {a } which coverges to a, the f(a coverges to f(a. Let s try to use this. First, we formulate the problem i the right way so we ca use the theorem. What should be our f(x? Note that a looks suspiciously like the fuctio e x. I fact, if f(x = e x, a = f( l= l 2. Great. We kow that f(x = e x is cotiuous o R. Now, ote that l= l 2 is the -th partial sum of a familiar sequece = 2. We already kow that this sequece of partial sums coverge: l= k! l 2 = = 2 = π2 6 Page 6 of 0
7 Therefore, we coclude that k=0 ( l= k l 2 k! ( π 2 = f = e π2 /6 6 Example 2.5. Show that the sequece, 2, 2 2, 2 2 2,... coverges. Proof. We will show that this sequece is mootoically icreasig ad bouded. First, let s show that the sequece is mootoically icreasig. Note that the sequece ca be recursively defied as follows: a =, a + = 2a Let s proceed by iductio. For =, a 2 = 2 = a. Now suppose a k a k+ for all k =,...,. We wat to show that a + a. We have: a 2 = ( 2a 2 = 2a 2a = ( 2a 2 = a 2 + Hece, the sequece is mootoically icreasig. Now, let s show that the sequece is bouded from above. How are we goig to do this? Iductio! We claim that the sequece is bouded above by 2. The base case is trivial: a = 2. Now suppose for all k =,...,, we have a k 2. The, ote that a + = 2a = 2 a 2 2 = 2 Therefore, a + 2. Hece, we coclude that a 2 for all. Sice {a } is a mootoically icreasig sequece which is bouded from above, we coclude that the sequece coverges. (I fact, a = Series Example 2.6. Show that the followig series coverges coditioally =2 ( log 2 Proof. First, let s show that the series coverges. To do this, we ca use the Alteratig Series Test. By the Alteratig Series Test, it suffices to show that log 2 = 0 Note that for all 2 we have the followig iequalities: 0 log 2 However, we have 0 = 0 ad = 0. Therefore, by the squeeze theorem we coclude log 2 = 0 Hece, the Alteratig Series Test tells us that above series coverges. Page 7 of 0
8 Now, we show that the series does ot coverge absolutely. I other words, we wat to show that the series log 2 =2 diverges. First, let s write out the first few terms of the series. =2 log 2 = 2 log log log log log Now, we observe the followig set of iequalities: 2 log log log log log log log log 2 2 = 2 = 2 (9 2 4 log 2 4 = = 2 2 (0 4 8 log 2 8 = = 2 3 ( We start to see a patter. More precisely, we see that (2 + log 2 ( log log 2 2 = 2 Now, we are ready to show that the series diverges absolutely. Cosider the partial sums S 2. We have S 2 = 2 log =2 2 ( = 2 log = 2 + ( ( 3 log log ( (2 + log 2 ( log 2 2 (2 (3 But ote that we have ow bouded the partial sum S 2 from below by half of the -th partial sum of the harmoic series. However, we already kow that the harmoic series diverges, which implies that the partial sums of the harmoic series is a diverget sequece. Therefore, by Compariso Test, we coclude that S 2 also diverges. Thus, we coclude that the series log 2 diverges. =2 Combiig the two results, we coclude that the series =2 ( log 2 coverges coditioally. Example 2.7. For N, let α( deote the umber of prime factors of. For example, α(2 =, α(4 = 2, α(6 = 2, α(20 = 5, etc. Determie whether the followig series coverges: = α( 3 Proof. Wheever we have a series ad we wat to get a sese of whether it coverges, it is a good idea to compare it to series that we kow already. I this particular example, we will compare it to. 2 First, ote that α( for all. Why? Suppose you have a umber with k prime factors. The smallest (4 (5 Page 8 of 0
9 such umber is 2 k > k. Usig this, we see that α( 3 3 = 2 Now, by the compariso test, we coclude that our series coverges. 2.4 Cotiuity ad Differetiability of Fuctios Example 2.8. Cosider f(x = { x 2 x Q 0 x Q Is f(x cotiuous ad/or differetiable at 0? Proof. We will show that f(x is ideed differetiable at 0. Sice differetiable fuctios are always cotiuous, it suffices to show that f(x is differetiable at 0. We will, however, also show that f(x is cotiuous at 0 to get more practice with ε δ proofs. Let ε > 0. The, we choose δ = ε. Now, let x be ay real umber such that x 0 < δ. The, f(x f(0 = f(x 0 = f(x = If x Q, the f(x 0 = 0 < ε so we are doe. If x Q, we have f(x = x 2 < δ 2 = ε { x 2 x Q 0 x Q Therefore, f(x = 0 = f(0 ad thus, f(x is cotiuous at 0. x 0 Now, we show that f(x is differetiable at 0. Let s look at the defiitio of the derivative. We say that f(x is differetiable at 0 if the it f(h f(0 h 0 h exists. Sice f(0 = 0, this is just f(h h 0 h But ote that because of the way f(x is defied, for all h, we have But ote that h 2 f(h h 2 h 2 h f(h h h 2 h h 2 h 0 h = h 0 h 2 h = 0 f(h Therefore, by the squeeze theorem, we coclude that h 0 h f (0 = 0. = 0. Hece, f(x is differetiable at 0 ad Example 2.9. Show that the fuctio { ( x si x x 0 f(x = 0 x = 0 is cotiuous but ot differetiable at 0. Page 9 of 0
10 Proof. Recall that the fuctio g(x = si ( x does ot have a it as x 0. Let s see why f(x is cotiuous at 0. To show that f(x is cotiuous at 0, we must show that for all ε > 0, we ca fid δ > 0 such that if x < δ, the f(x f(0 = f(x < ε. As always, we start by examiig the quatity that we wat to make small i.e. f(x. For ozero x, we have that ( f(x = x x si Sice si y, we ca boud this as follows ( f(x = x x si x But by choosig δ, we get to cotrol x. Let s start the proof. Let ε > 0 ad let δ = ε. Suppose x 0 = x < δ. The, f(x 0 = f(x x < δ = ε Therefore, we coclude that f(x = 0 = f(0. Hece, f(x is cotiuous at 0. x 0 Now, we show that f(x is ot differetiable at 0. I other words, we what to show that the followig it does ot exist: f(h f(0 f(h = h 0 h h 0 h = h 0 h si ( h = si h h 0 ( h However, we have already see that this it does ot exist. Thus, by the defiitio of differetiability, we coclude that f(x is ot differetiable at 0. Page 0 of 0
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